The given parabola $\rm x^2 - 4x - 3y + 13 = 0$ follows the following equation of parabola:

$$\rm (x - h)^2 = 4a (y-k)$$

where (h,k) represents the vertex of the parabola.

To find the coordinates of the vertex, we factorize the given equation of the parabola

$\rm x^2 - 4x - 3y + 13 = 0$

$\rm or, x^2 - 4x + 13 = 3y$

$\rm or, x^2 - 4x + (4 - 4) + 13 = 3y$

$\rm or, x^2 - 4x + 4 + 9 = 3y$

$\rm or, x^2 - 2 \cdot 2 \cdot x + 2^2 = 3y - 9$

$\rm or, (x - 2)^2 = 3 (y - 3)$

$\rm or, (x - 2)^2 = \frac{4}{4} \cdot 3 \cdot (y - 3)$

$\rm \therefore (x - 2)^2 = 4 \cdot \frac{3}{4} ( y - 3)$

Now, we compare this equation with the above equation to get the values of (h,k). So, $\rm (h, k) = (2,3)$.

Also, $\rm a = \frac{3}{4}$.

The coordinates of the focus are given by (h, k + a). So, the coordinates of the focus is

$\rm (h, k+a) = \left ( 2, 3 + \frac{3}{4} \right )$

$\rm = \left ( 2 , \frac{12 + 3}{4} \right )$

$\rm = \left (2, \frac{15}{4} \right )$

The equation of the axis of the parabola is given by

$\rm x= h$

$\rm \therefore x = 2$

The equation of the directrix of the parabola is given by

$\rm y = k - a$

$\rm or, y = 3 - \frac{3}{4}$

$\rm or, y = \frac{12 - 3}{4}$

$\rm or, y = \frac{9}{4}$

$\rm or, 4y = 9$

$\rm \therefore 4y - 9 = 0$ is the required equation of the directrix of the parabola.

The length of the latus rectum is given by $\rm 4a = 4 \cdot \frac{3}{4} = 3$ units.

We know, that the latus rectum passes through the focus of the parabola. For the given parabola, the y-coordinates of the focus are the same as the y-coordinates of the ends of the latus rectum.

Let us assume that the point P $\rm (x_1, k+a)$ and $\rm (x_2, k+a)$ are the coordinates of the ends of the latus rectum. We know, $\rm k + a = \frac{15}{4}$.

Put the value of $\rm y = k + a$ in the equation of the parabola to get the two values of x, we get,

$\rm (x - 2)^2 =3 (y - 3)$

$\rm or, (x - 2)^2 = 3 \left ( \frac{15}{4} - 3 \right )$

$\rm or, (x - 2)^2 = 3 \cdot \left ( \frac{15 - 12}{4} \right )$

$\rm or, (x - 2)^2 = 3 \cdot \frac{3}{4}$

$\rm or, (x - 2)^2 = \frac{9}{4}$

$\rm or, (x - 2) = \pm \sqrt{ \frac{9}{4} }$

$\rm or, x - 2 = \pm \frac{3}{2}$

Taking positive sign, we get,

$\rm or, x - 2 = \frac{3}{2}$

$\rm or, x = \frac{3}{2} + 2$

$\rm or, x = \frac{3 + 4}{2}$

$\rm \therefore x= \frac{7}{2}$

Let this value of x be the value of $\rm x_1$. So, $\rm x_1 = \frac{7}{2}$.

Taking negative sign, we get,

$\rm or, x - 2 = - \frac{3}{2}$

$\rm or, x = - \frac{3}{2} + 2$

$\rm or, x = - \frac{3 - 4}{2}$

$\rm or, x = \frac{1}{2}$

$\rm \therefore x = \frac{1}{2}$

Let this value of x be the value of $\rm x_2$. So, $\rm x_2 = \frac{1}{2}$.

Hence, the required coordinates of the ends of the latus rectum are $\rm \left ( \frac{7}{2}, \frac{15}{4} \right )$ and $\rm \left ( \frac{1}{2}, \frac{15}{4} \right )$